Why doubling the product fixes the splendid-number obstruction

laggyLucio·Apr 26, 2025(edited)

Two is splendid and four isn't, so both addition and ordinary multiplication fail immediately. I don't see why inserting a factor of 2 into the product should repair things. Problem: We call a positive integer

N

splendid, if

N = (a - b)^2 + (b - c)^2 + (c - a)^2

for some integers

a

b

and

c

. If

M

and

N

are splendid positive integers, is the sum

M + N

or the product

MN

also necessarily splendid? How about the product

2MN

? Source: Baltic Way 2020 · MNET-01G9

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5 replies

pixelated_toast·Apr 26, 2025(edited)

Translate c to zero and write

Q(x,y)=x^2-xy+y^2

. A splendid number has the form 2Q(x,y). This quadratic form has a multiplicative identity.

laggyLucio·Apr 27, 2025(edited)

Is this the norm from Eisenstein integers? I'd prefer an explicit identity I can verify without introducing the ring.

pixelated_toast·Apr 28, 2025(edited)

Use

Q(x,y)Q(u,v)=Q(xu-yv,xv+yu-yv)

. It follows by expansion, or by multiplying x+yω and u+vω with

ω^2+ω+1=0

snorlax_afk·Apr 28, 2025(edited)

If M=2Q(x,y) and N=2Q(u,v), then

2MN=8Q(A,B)=2Q(2A,2B)

, where A,B are those two expressions. Thus the triple (2A,2B,0) realizes 2MN.

laggyLucio·Apr 29, 2025(edited)

The scaling accounts for the extra factor. For completeness, 4 cannot be splendid because three integer squares summing to 4 would be 4,0,0, but two zero pairwise differences force the third to be zero too.

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